Multiple choice

Sum to $n$ terms of the series $ \displaystyle \frac {1}{1.2.3}+\frac {3}{2.3.4}+\frac {5}{3.4.5}+\frac {7}{4.5.6}+ ....$is

  1. $\displaystyle \frac {n(n+1)}{2(n+2)(n+3)}$
  2. $\displaystyle \frac {n(3n+5)}{4(n+1)(n+2)}$
  3. $\displaystyle \frac {1}{6}-\frac {5}{(n+1)(n+4)}$
  4. $\displaystyle \frac {n(3n+1)}{4(n+1)(n+2)}$
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D Correct answer
Explanation

The general term is a_n = (2n-1) / (n(n+1)(n+2)). Using partial fractions, this simplifies to 1/2 * [1/(n(n+1)) - 1/((n+1)(n+2))]. Summing this telescoping series leads to the result n(3n+1) / (4(n+1)(n+2)).

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